- Consider a

**circular ring of radius 'a' which carries a uniform line charge density ρ**as shown in figure.

_{L}- We need to find out

**electric field**at a point P (0, 0, h) on the z axis (z > 0).

- Electric field intensity

**(E) due to any line charge (ρ**) in general is given as:

_{L}In this case,

**dl = a dφ**(Since the differential part

**dl is a differential arc**)

**R**

^{2}= a^{2}+ h^{2}Consider the triangle shown in the above figure

**a a**

_{ρ}+ R = h a_{z}→ R = - a a_{ρ}+ h a_{z}**a**

_{R}= R / | R |**a**

_{R}= - a a_{ρ}+ h a_{z}/ ( a^{2}+ h^{2})^{1/2}Substituting all these values in the above equations,

**the electric field intensity E**becomes:

- For every element

**dl there is a corresponding element diametrically opposite**that gives an equal but opposite

**dE**so that

_{ρ}**the two contributions cancel each other.**

Hence

**contribution along a**due to symmetry

_{ρ}**adds up to zero.**

- Therefore the final

**electric field intensity**

**at point (0, 0, h) has only z component.**

- Hence Electric field intensity (E) due to a

**circular ring (of radius a carrying a uniform charge ρ**placed on the

_{L})**x-y plane**and if the

**point of interest is any point on z axis,**then it is given as:

- Similarly

**if the ring is placed on x-z plane and point of interest is any point on y- axis,**then E is given as:

**ALSO READ:**

**- Introduction To Electrostatics.**

**- Coulomb's law.**

**- Electric Field Intensity (E).**

**- Electric Lines Of Forces /Streamlines / Electric Flux (ψ) .**

**- Electric Flux Density (D).**

**- Electric Field Intensity Due To a Finite and Infinite Line Charge.**

**- Electric Field Intensity Due To a Infinite Sheet Charge.**

**- Electric Field Intensity Due To a Circular Ring Charge.**

**- Electric Field Intensity Due To a Circular Disk Charge.**

**- Numericals / Solved Examples - Electric Force and Field Intensity.**

**- Numericals / Solved Examples - Electric Field Intensity - Line, Surface and Mixed Charge Configuration.**

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